Problem 47 - Chemistry and Chemical Reactivity

Question

At \(450^{\circ} \mathrm{C}, 3.60\) mol of ammonia is placed in a \(2.00-\mathrm{L}\) vessel and allowed to decompose to the elements. $$ 2 \mathrm{NH}_{3}(\mathrm{g}) \rightleftarrows \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) $$ If the experimental value of \(K\) is 6.3 for this reaction at this temperature, calculate the equilibrium concentration of each reagent. What is the total pressure in the flask?

Step-by-Step Solution

Answer

Equilibrium concentrations: \([\mathrm{NH}_3] = 1.80 - 2x, [\mathrm{N}_2] = x, [\mathrm{H}_2] = 3x\). Total pressure: calculated using ideal gas law.

Step 1: Calculate Initial Concentrations

The initial amount of NH₃ is 3.60 mol in a 2.00 L container. Therefore, the initial concentration of NH₃ is given by \[ [\text{NH}_3]_0 = \frac{3.60}{2.00} \text{ mol/L} = 1.80 \text{ M} \]Initially, the concentrations of N₂ and H₂ are 0 M because only NH₃ is added.

Step 2: Define the Change in Concentrations

Let the change in concentration of NH₃ at equilibrium be -2x, then \[ [\text{N}_2] = x \text{ and } [\text{H}_2] = 3x \] at equilibrium because the stoichiometry of the balanced equation is 2:1:3.

Step 3: Setup the Equilibrium Expression

The equilibrium expression for the reaction is \[ K = \frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2} \] and substitute known values: \[ K = 6.3 = \frac{(x)(3x)^3}{(1.80 - 2x)^2} \]

Step 4: Solve for x

Solve the equation \[ 6.3 = \frac{(x)(27x^3)}{(1.80 - 2x)^2} \]By simplifying and rearranging terms, we obtain a fourth-degree polynomial in x, which can be solved to find the value of x. This might require numerical methods or approximations.

Step 5: Calculate Equilibrium Concentrations

Once the value of x is known, substitute it back into the expressions for equilibrium concentrations:\[ [\text{NH}_3] = 1.80 - 2x, \quad [\text{N}_2] = x, \quad [\text{H}_2] = 3x \] These are the equilibrium concentrations for NH₃, N₂, and H₂ respectively.

Step 6: Calculate Total Pressure

The total pressure can be found using the ideal gas law applied to the total moles of gas at equilibrium:\[ P = \frac{nRT}{V} \]where R is the ideal gas constant, T is the temperature in Kelvin, and V is the volume. Convert the concentration of each gas to moles using the volume of the flask and sum them to obtain total moles.

Key Concepts

Ammonia Decomposition

In the context of chemical equilibrium, ammonia decomposition is the process where ammonia (\(\text{NH}_3\),) breaks down into its elemental components,\(\text{nitrogen (N}_2)\) and \(\text{hydrogen (H}_2)\). This is represented by the balanced chemical equation:\[2 \text{NH}_3 (\text{g}) \rightleftarrows \text{N}_2 (\text{g}) + 3 \text{H}_2 (\text{g})\]Here, you can see that two moles of ammonia decompose to create one mole of nitrogen and three moles of hydrogen. Understanding this stoichiometry is crucial because it helps us manage the changes in concentrations of the reactants and products over time. This reversible reaction means it can proceed in both forward and backward directions until it reaches a state of equilibrium, where the rate of decomposition of ammonia equals the rate of its formation from nitrogen and hydrogen. The behavior of gases like ammonia during this decomposition depends on key factors such as temperature and concentration.

Equilibrium Constant

The equilibrium constant, represented as \(K\), provides insight into the proportions of products and reactants at equilibrium for a reversible reaction like the ammonia decomposition. In this scenario, the equilibrium constant value is given as 6.3 at \(450^{\circ} \text{C}\). This value is crucial as it tells us the likelihood of the reaction proceeding in the forward direction compared to the reverse.The equilibrium expression for our reaction is:\[K = \frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2}\]In simpler terms, it is a ratio that shows the relationship between the concentration of the products and the concentration of the reactants when the reaction has reached equilibrium. A larger \(K\) value implies a greater concentration of products at equilibrium, which indicates that the reaction favors the formation of products. Calculating and understanding this key value is essential when assessing how far a reaction will proceed under given conditions.

Stoichiometry

Stoichiometry involves the study of quantitative relationships in chemical reactions. In the decomposition of ammonia, it allows us to predict how much \(\text{NH}_3\) will decompose into \(\text{N}_2\) and \(\text{H}_2\) at equilibrium.Given the balanced equation:\[2 \text{NH}_3 \rightleftarrows \text{N}_2 + 3 \text{H}_2\]The stoichiometry tells us that:

• For every 2 moles of \(\text{NH}_3\) that decompose, 1 mole of \(\text{N}_2\) and 3 moles of \(\text{H}_2\) are produced.
• This relationship helps us set up a conversion factor, where the change in concentration or moles of each substance is proportional to their coefficients in the balanced equation.

This stoichiometric relationship is integral when calculating the concentrations of all species at equilibrium, in conjunction with the equilibrium constant. It simplifies complex chemical concepts by connecting the abstract world of molecules to real-world measurable quantities like moles and liters.

Ideal Gas Law

The Ideal Gas Law is a handy equation that provides a relationship between the physical properties of gases: pressure \(P\), volume \(V\), temperature \(T\), and the number of moles \(n\). It is expressed as:\[P = \frac{nRT}{V}\]where \(R\) is the ideal gas constant.In the ammonia decomposition exercise, we use the Ideal Gas Law to compute the total pressure in the reaction vessel once equilibrium is established. Here's how:

• Calculate the total number of moles of gas present at equilibrium.
• Apply the Ideal Gas Law using the values for \(R\) (ideal gas constant), \(T\) (temperature in Kelvin), and \(V\) (volume of the vessel).

This law is particularly useful because it provides a way to directly connect the "macroscopic" quantities of pressure and volume with the chemical "microscopic" quantities of moles and temperature, making it a fundamental tool in solving equilibrium problems involving gases.